Question

若(ax05-2xy+y05)-(-x05+bxy+2y05)=5x05-9xy+cy05成立,求a,b,c的值

Answer 1

∵(ax05-2xy+y05)-(-x05+bxy+2y05)=5x05-9xy+cy05

∴(a+1)x05-(b+2)xy-y05=5x05-9xy+cy05成立

∴同类项的系数相等

∴a+1=5 ,a=4

b+2=9,b=7

c=-1

∴a=4,b=7,c=-1